Question 1

Consider the following three statements about link state and distance vector routing protocols, for a large network with 500 network nodes and 4000 links
[S1] The computational overhead in link state protocols is higher than in distance vector protocols.
[S2] A distance vector protocol (with split horizon) avoids persistent routing loops, but not a link state protocol.
[S3] After a topology change, a link state protocol will converge faster than a distance vector protocol.
Which one of the following is correct about S1, S2, and S3?

A : S1, S2, and S3 are all true
B : S1, S2, and S3 are all false.
C : S1 and S2 are true, but S3 is false
D : S1 and S3 are true, but S2 is false.


     Correct answer is :D

     Solution :
      The DV rely on the info from directly connected neighbours in order to calculate and accumulate route information. DV require very little overhead as compared to LS. LS uses data from whole link for generating the routing table so huge overhead is there.
    S3 is also true as Distance Vector protocol has count to infinity problem and converges slower.
    S2 is false. In distance vector protocol, split horizon with poison reverse reduces the chance of forming loops and uses a maximum number of hops to counter the 'count-to-infinity' problem. These measures avoid the formation of routing loops in some, but not all, cases

  •   Question 2

    Which one of the following are used to generate a message digest by the network security protocols?
    (P) RSA
    (Q)SHA - I
    (R) DES
    (S) MD5

    A : P and R only
    B : Q and R only
    C : Q and S only
    D : R and S only


     Correct answer is :C

     Solution :
      RSA and DES are for Encryption where MD5 and SHA – 1 are used to generate Message Digest.

  •   Question 3

    Identify the correct order in which the following actions take place in an interaction between a web browser and a web server.
    1. The web browser requests a webpage using HTTP.
    2. The web browser establishes a TCP connection with the web server.
    3. The web server sends the requested webpage using HTTP.
    4. The web browser resolves the domain name using DNS.

    A : 4,2,1,3
    B : 1,2,3,4
    C : 4,1,2,3
    D : 2,4,1,3


     Correct answer is :A

     Solution :
      First of all the browser must now know what IP to connect to. For this purpose browser takes help of Domain name system (DNS) servers which are used for resolving hostnames to IP addresses. As browser is an HTTP client and as HTTP is based on the TCP/IP protocols, first it establishes a TCP connection with the web server and requests a webpage using HTTP, and then the web server sends the requested webpage using HTTP. Hence the order is 4,2,1,3

  •   Question 4

    Consider a token ring network with a length of 2km having 10 stations including a monitoring station. The propagation speed of the signal is 2 x 108 m/s and the token transmission time is ignored. If each station is allowed to hold the token for 2 microsec, the minimum time for which the monitoring station should wait (in microsec) before assuming that the token is lost is _______.


     Correct answer is :30

     Solution :
      Given Length (d) = 2 Km
    No. of Stations (m) = 10 Propagation Speed (v) = 2 × 108 m/s
    THT = 2?s
    So, Max. TRT = TP in the Ring + No. of Active Stations * THT
    =10 × 10-6 + 10 × 2 × 10-6
    =30 s

  •   Question 5

    Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2kB. The time taken (in msec) by the TCP connection to get back to 32KB congestion window is ______


     Correct answer is :1100 to 1300

     Solution :
      Given that at the time of Time Out, Congestion Window Size is 32KB and RTT = 100ms When Time Out occurs, for the next round of Slow Start, Threshold = (size of Cwnd) / 2 It means Threshold = 16KB
    Slow Start 2KB
    16KB ----------- Threshold reaches. So Additive Increase Starts
    So, Total no. of RTTs = 11

  •   Question 6

    Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________.


     Correct answer is :5

     Solution :
      Given L =1KB
    B= 1.5Mbps
    Tp =50ms
    η = 60%
    Efficiency formula for SR protocol is
    η = W/(1+2a)= 60/100
    Tx= L/B = 8000/1.5 * 106 = 5.3 ms
    a= Tp/Tx =50/5.3 = 500/53 = 9.43
    =60/100 = W/19.86 => W= 11.9 ≅ 12
    W = 2n-1 = 12 => 2n = 24 ≅ = 25 => n=5

  •   Question 7

    A 4-way set-associative cache memory unit with a capacity of 16 KB is built using a block size of 8 words. The word length is 32 bits. The size of the physical address space is 4 GB. The number of bits for the TAG field is _____


     Correct answer is :20

     Solution :
      tag = 32 - (7+5) =20 bits

  •   Question 8

    Which one of the following is TRUE about the interior gateway routing protocols – Routing Information Protocol (RIP) and Open Shortest Path First (OSPF)?

    A : RIP uses distance vector routing and OSPF uses link state routing
    B : OSPF uses distance vector routing and RIP uses link state routing
    C : Both RIP and OSPF use link state routing
    D : Both RIP and OSPF use distance vector routing


     Correct answer is :A

     Solution :
      RIP Uses Distance Vector Routing and OSPF uses Link State Routing.

  •   Question 9

    Which one of the following socket API functions converts an unconnected active TCP socket into a passive socket?

    A : connect
    B : bind
    C : listen
    D : accept


     Correct answer is :C

     Solution :
      (a) The connect function is used by a TCP client to establish a connection with a TCP server.
    (b) The bind function assigns a local protocol address to a socket. With the Internet protocols, the protocol address is the combination of either a 32-bit IPv4 address or a 128-bit IPv6 address, along with a 16-bit TCP or UDP port number.
    (c) The listen function converts an unconnected socket into a passive socket, indicating that the kernel should accept incoming connection requests directed to it. d) Accepts the connection request after listening.

  •   Question 10

    In the diagram shown below, L1 is an Ethernet LAN and L2 is a Token-Ring LAN. An IP packet originates from sender S and traverses to R, as shown. The links within each ISP and across the two ISPs, are all point-to-point’ optical links. The initial value of the TTL field is 32. The maximum possible value of the TTL field when R receives the datagram is ____________.


     Correct answer is :26

     Solution :
      The TTL field is set by the sender of the datagram, and reduced by every router on the route to its destination. So, there are 5visits at 5 routers and one visit at receiver R in above figure which leads 32 – 6 = 26.

  •   Question 11

    Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 106 bytes / sec. A user on host A sends a file of size 103 bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these

    A : T1 < T2 < T3
    B : T1 > T2 > T3
    C : T2 = T3, T3 < T1
    D : T1 = T3, T3 > T2


     Correct answer is :D

  •   Question 12

    An IP machine Q has a path to another IP machine H via three IP routers R1, R2, and R3.
    H acts as an HTTP server, and Q connects to H via HTTP and downloads a file. Session layer encryption is used, with DES as the shared key encryption protocol. Consider the following four pieces of information:
    [I1] The URL of the file downloaded by Q
    [I2] The TCP port numbers at Q and H
    [I3] The IP addresses of Q and H
    [I4] The link layer addresses of Q and H
    Which of I1, I2, I3, and I4 can an intruder learn through sniffing at R2 alone?

    A : Only I1 and I2
    B : Only I1
    C : Only I2 and I3
    D : Only I3


     Correct answer is :C

     Solution :
      An Intruder can’t learn [I1] through sniffing at R2 because URLs and Download are functioned at Application layer of OSI Model.
    An Intruder can learn [I2] through sniffing at R2 because Port Numbers are encapsulated in the payload field of IP Datagram. An Intruder can learn
    [I3] through sniffing at R2 because IP Addresses and Routers are functioned at network layer of OSI Model.
    An Intruder can’t learn [I4] through sniffing at R2 because it is related to Data Link Layer of OSI Model.

  •   Question 13

    A graphical HTML browser resident at a network client machine Q accesses a static HTML webpage from a HTTP server S. The static HTML page has exactly one static embedded image which is also at S. Assuming no caching, which one of the following is correct about the HTML webpage loading (including the embedded image)?

    A : Q needs to send at least 2 HTTP requests to S, each necessarily in a separate TCP connection to server S
    B : Q needs to send at least 2 HTTP requests to S, but a single TCP connection to server S is sufficient
    C : A single HTTP request from Q to S is sufficient, and a single TCP connection between Q and S is necessary for this
    D : A single HTTP request from Q to S is sufficient, and this is possible without any TCP connection between Q and S


     Correct answer is :B

     Solution :
      Whenever a browser opens a webpage, it makes a separate request for each object of page like image, css, javascript, etc. However if multiple resources are served from same server, then one TCP connect is sufficient. There can be more than one TCP connections also but the it can be done in one connection so by condition of sufficiency it is true.

  •   Question 14

    In the following pairs of OSI protocol layer/sub-layer and its functionality, the INCORRECT pair is

    A : Network layer and Routing
    B : Data Link Layer and Bit synchronization
    C : Transport layer and End-to-end process communication
    D : Medium Access Control sub-layer and Channel sharing


     Correct answer is :B

     Solution :
      (a) One of the main functionality of Network Layer is Routing. So Option (a) is CORRECT.
    (b) Bit Synchronization is always handled by Physical Layer of OSI model but not Data Link Layer. So Option (b) is INCORRECT.
    (c) End – to – End Process Communication is handled by Transport Layer. So Option (c) is CORRECT.
    (d) MAC sub layer have 3 types of protocols (Random, Controlled and Channelized Access).

  •   Question 15

    A bit-stuffing based framing protocol uses an 8-bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is

    A : 0111110100
    B : 0111110101
    C : 0111111101
    D : 0111111111


     Correct answer is :B

     Solution :
      Given 8 – bit delimiter pattern of 01111110.
    Output Bit string after stuffing is 01111100101
    Now, Input String is 0111110101

  •   Question 16

    An IP router implementing Classless Inter-domain routing (CIDR) receives a packet with address The router’s routing table has the following entries. The identifier of the output interface on which this packet will be forwarded is _____.


     Correct answer is :1

     Solution :
      Given address to the first field of given routing table
    131.0001 0111.151.76
    131.0001 0000.0.0 ( given mask bits = 12) Matched
    Coming to the 2nd field of given Routing table
    131.0001 0111.151.76
    131.0001 0100.0.0 ( given mask bits = 14) Not matched.
    Coming to the 3rd field of given Routing table
    Error! Not a valid link.
    131.0001 0111.151.76
    131.0001 0

  •   Question 17

    Every host in an IPv4 network has a 1-second resolution real-time clock with battery backup. Each host needs to generate up to 1000 unique identifiers per second. Assume that each host has a globally unique IPv4 address. Design a 50-bit globally unique ID for this purpose. After what period (in seconds) will the identifiers generated by a host wrap around?


     Correct answer is :256

     Solution :
      Given that each host has a globally unique IPv4 Address and we have to design 50 – bit unique Id. So, 50 – bit in the sense (32 + 18). So, It is clearly showing that IP Address (32 – bit) followed by 18 bits.
    1000 unique Ids => 1Sec
    218 / 1000 = 28 = 256

  •   Question 18

    An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are

    A : MF bit: 0, Datagram Length: 1444; Offset: 370
    B : MF bit: 1, Datagram Length: 1424; Offset: 185
    C : MF bit: 1, Datagram Length: 1500; Offset: 370
    D : MF bit: 0, Datagram Length: 1424; Offset: 2960


     Correct answer is :A

    TOTAL = 18
    CORRECT / TOTAL = /18